The equation of a circle $C$ is $x^2+y^2-4x+8y-16 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-4x) + (y^2+8y) = 16$ $(x^2-4x+4) + (y^2+8y+16) = 16 + 4 + 16$ $(x-2)^{2} + (y+4)^{2} = 36 = 6^2$ Thus, $(h, k) = (2, -4)$ and $r = 6$.